Formula for concentration of force (battle axe vs. a mace)

[Duke Patrick]

I really hope someone can help. I have spent the last week several hours a day trying to get this answer via Google.

I am doing research on the practical difference between a mace and axe in combat. Strangely enough there's almost nothing concerning the techniques of using a mace in period manuscripts. At least not any translated to english.

I have most of my answers from physical testing and my 30 years of real life tournament fighting. But the one problem I am having is trying to get a straight forward answer. A formula to find the concentration of force from a blunt "hammer" head of 1x1 inches to a blade 0.1 x 2 inches (not sharp as melee weapons get dull fast in armored combat) .

What I really want to know is how much more deformation will occur to a plate of steel from the mace to the axe. A "modifier" to be use on a mace that say does 30 pts deformation damage but that same mace reshaped into an axe (same ballance, length, weight etcetera) would do X amount more of deformation damage.

It can be simplified as this is for programming a game not anything important and needs to be written out like 2.0 x (weight x 0.5) / sqr) or similar to that.

Thanks and sorry for such a frivolous question, but I am stumped on this other than to think it is just a logical answer like factor = (1.0 x 1.0 ) / (0.1 x 2.0) .

I feel it cannot be that easy based on the other formulas I have researched such as the fact that the length of a hammer handle is a quadrilateral factor not a liner one.

 

[Dale]

I think that your "concentration of force" is just pressure, which is force/area. Deformation is related to pressure and stiffness by: $$\frac{\text{change in length}}{\text{length}}=\frac{\text{pressure}}{\text{stiffness}}$$

 

[rumborak]

I think maybe one simplification that could give you a formula would be treating it as if the mace/axe is hitting a very viscous material, and that it is eventually stopped by drag forces inside the material. So, the axe would have less drag because of its smaller surface, thus traveling further into the material.

 

[jack action]

I would go with @Dale and say pressure as well. Since the force applied is the same for both weapons, only the area changes. So:
$$P_{mace} =\frac{F}{A_{mace}}$$
$$P_{axe} =\frac{F}{A_{axe}}$$
Thus:
$$\frac{P_{mace}}{P_{axe}} =\frac{A_{axe}}{A_{mace}} = \frac{0.1 \times 2}{1 \times 1}$$
Just like you intuitively guessed.

 

[Dr.D]

This looks to me like a Hertz contact stress problem. Looked at in that manner, everything depends upon the radii of curvature of the two bodies at the point of contact (along with Young's modulii). You might want to research Hertz contact stress to see how this is calculated.

 

[Nidum]

This is a kinetic response and plate deformation problem . What happens in a particular situation depends on many things .

If the problem is relating to impact damage to a relatively thin steel plate then much depends on how the plate is supported . In an old type 'suit of armour' for example the plate might be supported on thick padding in turn supported by the wearers body .

Where the blow is (relatively) low energy and there is no actual through penetration the impact damage can be modeled reasonably accurately using standard plate deformation theory .

 

[Duke Patrick]

Sorry I guess I was not clear enough. ALL other factors are the same.

I need a simple (program script ready ) formula that tells me an approximate difference in the concentration of force from a 1x1 face of the hammer (mace) to a "dull" 0.1 x 2.0 edge of the axe. Again with the axe and hammer being exactly the same weight, length, and balance (along the handle) and the target is the same and supported the same.

Jack Action's answer is the kind of answer I hoping for. And that answer seems logical to me. However I thought I would check to see if it was an unintuitive answer similar to the quadrilateral formula used for the length of a hammer handle.

Here is my dilemma with Jack Action's answer: That would mean the axe would be doing about 5 times the damage. In real life testing it is definitely is not 5 times the damage. It is closer to about 25 to 100 percent more. So I am missing some other factor.

Thank you all for your help and your time.

 

[jack action]

That would mean the axe would be doing about 5 times the damage. In real life testing it is definitely is not 5 times the damage. It is closer to about 25 to 100 percent more.

How do you define «damage»?

 

[Duke Patrick]

By damage I mean deformation of the plate.

Ok, I went here: http://www.amesweb.info/HertzianContact/HertzianContact.aspx
(Thank you so very much Dr.D)

Using 1.0 inch strike area for the mace and (0.1 x 2.0 = 0.2) inch strike area for the dull blade of an axe…

Mace: 1 inch for "diameter of object" returned Max Shear Stress of 96.7

then

Axe: 0.2 inch for "diameter of object" returned Max Shear Stress of 214

ALL other factors I left as default on the calculator as those factors would be the same form this hypothetical mace to the axe.

And other results such as "depth of shear stress" and "contact pressure" were all the same, about TWO times from the mace to the axe. That matches my real life experience a heck of a lot more. But am I miss using the Hertzian Contact concept here way too much to be useful?EDIT:
OK, I think I found out the esoteric factor that is making this confusing, from a book on stress testing I read that the effect of the Hertzian contact stress dissipates with the distance squared from the point of contact.

So I THINK that even though the axe blade is 10 times more force at the areas of contact, the total force “result” is reduced because of the small footprint.

A little like how friction and area of contact works…

However then there is the separate but important bending deformation at the points of support. :(
Wow how do you engineers and scientists keep all this juggling in your mind? This is another example on how two equally smart and educated people can disagree on physics. Either would be right or wrong depending on that one esoteric factor they both are forgetting.

I have seen this happen when my organization's safety committee was debating on if higher or lower pressure (softer or harder) air filled bladders were better for our throwing spears. The debate was that softer should be better but field tests proved otherwise and some speculated it was because of it bouncing off the target better with higher pressure others said that was silly... It turned out everyone was wrong, it was because the softer heads would allow the hard shaft to compress to the point that that it would "bottom out" effectively allowing the shaft to hit the opponent. That one key factor was overlooked in the sea of math they were debating.

 

[jack action]

But am I miss using the Hertzian Contact concept here way too much to be useful?

Yes. You are using an area in an equation that requires a diameter.

I tried something with the contact mechanics concept with a flat-end cylinder. The depth of the deformation is ##d=\frac{F}{2Ea}## where ##a## is the radius of the cylinder. Comparing the depth between 2 cylinders yields (assuming ##F## and ##E## are the same for both):
$$\frac{d_1}{d_2}= \frac{\frac{F}{2Ea_1}}{\frac{F}{2Ea_2}} = \frac{a_2}{a_1} = \sqrt{\frac{A_2}{A_1}}$$
Where ##A## is the area of the flat end of the cylinder. If we assume your weapon are cylinders, then the axe mark will be 2.236 times deeper than the one from the mace.

Bu if we want to compare the volume displaced, then we have to multiply the depth by the cylinder end's area ##\pi a^2##. The comparison will yield:
$$\frac{V_1}{V_2} = \frac{\pi a_1^2\frac{F}{2Ea_1}}{\pi a_2^2\frac{F}{2Ea_2}} = \frac{a_1}{a_2} = \sqrt{\frac{A_1}{A_2}}$$
Which is the inverse of the depth ratio. So for your weapons, the axe would displace 0.447 the volume of the mace.

I don't know if it helps or adds more to the confusion.

 

[Duke Patrick]

Yes. You are using an area in an equation that requires a diameter.
If we assume your weapon are cylinders, then the axe mark will be 2.236 times deeper than the one from the mace.

Thank you very very much Jack Action! This was what I needed! This place is a very cool part of the internet.

So this was using a diameter of 0.1 (radius 0.05) inch for the axe and 1.0 (radius 0.6) inch for the mace ?

 

[jack action]

So this was using a diameter of 0.1 (radius 0.05) inch for the axe and 1.0 (radius 0.6) inch for the mace ?

No, it was using the actual area of the axe and mace. It was assumed that they had a round shape while keeping the same area. The radius ##a## of an equivalent circle of the same area ##A## is ##a = \sqrt{\frac{A}{\pi}}##. Thus for the mace ##a## = 0.564" and for the axe ##a## = 0.252".

 

[Duke Patrick]

No, it was using the actual area of the axe and mace. It was assumed that they had a round shape while keeping the same area. The radius ##a## of an equivalent circle of the same area ##A## is ##a = \sqrt{\frac{A}{\pi}}##. Thus for the mace ##a## = 0.564" and for the axe ##a## = 0.252".

GOT IT! Makes perfect sense to me now. The force is the same so we only are concerned with the area of contact. Thank you so much.

 

[Nidum]

What I really want to know is how much more deformation will occur to a plate of steel from the mace to the axe

The indentation depth is only going to be a small contribution to the total plate deformation .

 

[sophiecentaur]

Loads of good ideas in this thread. It puts me in mind of the many threads about vehicle impacts. Sadly for the OP, the conclusions of those threads is that it' s too complex a problem for Physics theory alone. I think this calls for a large base of experimental results, specifically involving typical weapons, armor and users.
I guess Health and Safety hardly comes into it for these scamps! ;-)